Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.HashMap;
import java.util.Hashtable;
import java.util.List;
import java.util.Map;
import java.util.Scanner;

class Main
{
    double dp[][];
    double prob[];
    public static void main(String args[]) {
        Scanner sc = new Scanner(System.in);
        int coins=sc.nextInt();
        sc.nextLine();
        Main object=new Main();
        object.prob=new double[coins];
        for(int i=0;i<coins;i++)
        {
          object.prob[i]=sc.nextDouble();
        }
        object.dp=new double[coins+1][coins+1];
        object.dp[0][0]=1;
        for(int i=1;i<=coins;i++)
        {
          for(int j=0;j<=i;j++)
          {
            if(j==0)
              object.dp[i][j]=object.dp[i-1][j]*(1-object.prob[i-1]);
            else
              object.dp[i][j]=object.dp[i-1][j-1]*object.prob[i-1]+object.dp[i-1][j]*(1-object.prob[i-1]);
          }
        }
        double res=0;
        for(int i=coins/2+1;i<=coins;i++)
        {
               res=res+object.dp[coins][i];
        }
        System.out.println(res);
        sc.close();
     }
}